Chapter 24 - Chapter 25 2 Picture the Problem A person on...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
2. Picture the Problem : A person on the positive y -axis observes an electromagnetic wave that radiates from an electric charge that oscillates sinusoidally about the origin and along the x -axis. Strategy: In order for the radiation to reach the person, it must propagate in the positive y direction (toward the person). The electric field will oscillate in the same direction as the charge. The magnetic field will oscillate perpendicular to the electric field and perpendicular to the direction of propagation. Solution: 1. (a) The electric field will oscillate in the x direction . 2. (b) The magnetic field will oscillate in the z direction 3. (c) The electromagnetic wave will propagate in the positive y direction Insight: If the person were standing on the z- axis, and at some instant the electric field were in the + x direction, the magnetic field would be in the + y direction, as the wave propagated in the positive z direction. 6. Picture the Problem : A wave propagates in the positive z direction and has a known electric field vector. Strategy: Since the magnetic field is perpendicular to the electric field, the magnitudes of the x and y components of the magnetic field are proportional to the y and x components of the electric field, respectively, with the constant of proportionality given by equation 25-9. Use the Right-Hand Rule to determine the sign of each component. Solution: 1. (a) Since is perpendicular to the direction of propagation, the z component is zero. 2. (b) Switch the components of the electric field and divide by c to calculate 3. Insert the electric field components: Insight: Calculating the magnitude of the magnetic field gives as stated in the problem. This problem can also be solved by using the vector cross product (see Appendix A). 46. Picture the Problem : Two electromagnetic waves have different intensities. One wave has a maximum electric field of 52 V/m, and the second wave has a maximum magnetic field of 1.5 μ T. Strategy: Use equation 25-9 to calculate the electric field for wave 2. Compare the electric fields to determine which wave has the greater intensity. Then calculate the intensity of each wave using equation 25-10. Solution: 1. (a) Calculate the electric field of wave 2: 2. Since the intensity is proportional to the square of the electric field and wave 2 has the greater electric field, wave 2 has the greater intensity. 3. (b)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/03/2008 for the course PHYS 104 taught by Professor Dr.g during the Spring '08 term at Gwinnett Technical College.

Page1 / 5

Chapter 24 - Chapter 25 2 Picture the Problem A person on...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online